Question

AB||CD||EF. if angle ABC =70° and angle CEF=150°. then find angle BCE. ​

Answer 1

Answer:

Given:-

ABllCDllEF

$\green {\angle \: abc = \angle \: bcd } \\ (alternate \: \angle \: s) \\ \boxed{ \angle \: bcd = 70 \degree}$

$\red{ \angle \: fec + \angle \: ecd = 180 \degree} \\ (co \:.incident \angle \: s) \\ \boxed{\angle \: ecd = 180 - 150 = 30 \degree}$

Now;

Angle BCD= x+ Angle Ecd

$\blue{70 \degree = x + 30 \degree} \\ \boxed {x = 40 \degree}$

So, x= 40°

$\orange{ \mathfrak{plz \: mark \: brainliest}} \\ \pink{\tt{plz \: give \: thanks}}$

Answer 2

Step-by-step explanation:

1. <abc=<bcd. ( alternate interior angles )
2. <fec + <ecd = 180 ( co - interior angles are supplimentary ) ie . < ecd = 180 - 150 = 30
3. < bcd = 70 = x+ 30
4. :. x = 40 is the solution